Problem: Glen Davis is shooting free throws. Making or missing free throws doesn't change the probability that he will make his next one, and he makes his free throws $74\%$ of the time. What is the probability of Glen Davis making none of his next 4 free throw attempts? Choose 1 answer: Choose 1 answer: (Choice A) A ${4 \cdot0.74}$ (Choice B) B ${0.74^4}$ (Choice C) C ${4 \cdot(1 - 0.74)}$ (Choice D) D ${(1 - 0.74)^4}$
Answer: We know that ${26 \%}$ of the time, he'll miss his first shot $(100 \% - 74 \% = 26 \%)$. $26\%$ Then $26 \%$ of the time he misses his first shot, he will also miss his second shot, and $74 \%$ of the time he misses his first shot, he will make his second shot. $26\%$ Notice how we can completely ignore the rightmost section of the line now, because those were the times that he made the first free throw, and we only care about if he misses the first and the second. So the chance of missing two free throws in a row is $26\%$ of the times that he missed the first shot, which happens $26\%$ of the time in general. This is $26\% \cdot 26\%$, or $0.26 \cdot 0.26 \approx 0.068$. We can repeat this process again to get the probability of missing three free throws in a row. We simply take $26\%$ of probability that he misses two in a row, which we know from the previous step is $0.068 \approx {7\%}$. $26\%$ $7\%$ $26\%$ of ${7\%}$ is $0.26 \cdot 0.068 \approx 0.018$, or about ${2\%}$ : $26\%$ $7\%$ There is a pattern here: the chance of missing two free throws in a row was $0.26 \cdot 0.26$, and the probability of missing three in a row was $0.26 \cdot 0.068 = 0.26 \cdot (0.26 \cdot 0.26) = 0.26^3$. In general, you can continue in this way to find the probability of missing any number of shots. The probability of missing $4$ free throws in a row is $0.26 ^ 4 = (1 - 0.74)^4$.